JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2023
- Hard
- Numerical
A simple pendulum with length \(100\,cm\) and bob of mass \(250\,g\) is executing S.H.M. of amplitude \(10\,cm\). The maximum tension in the string is found to be \(\frac{x}{40}\,N\). The value of \(x\) is \(..........\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(99\)
Step-by-step Solution
Detailed explanation
\(\sin \theta_0=\frac{A}{l}=\frac{10}{100}=\frac{1}{10}\)
From conservation of energy
\(\frac{1}{2} m v^2=m g l(1-\cos \theta)\)
Maximum tension occurs at mean position.
\(\therefore T-m g=\frac{m v^2}{l}\)
\(\Rightarrow T=m g+\frac{m v^2}{l}\)
\(\therefore T=m g+2 m g(1-\cos \theta)\)
\(=m g\left[1+2\left(1-\sqrt{1-\sin ^2 \theta}\right)\right]\)
\(=m g\left[3-2 \sqrt{1-\frac{1}{100}}\right]\)
\(=\frac{250}{1000} \times 9.8\left[3-2\left(1-\frac{1}{200}\right)\right]=\frac{99}{40}\)
\(\therefore x=99\)
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